Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 2}{x + 2} = \dfrac{-9x - 12}{x + 2}$
Answer: Multiply both sides by $x + 2$ $ \dfrac{x^2 + 2}{x + 2} (x + 2) = \dfrac{-9x - 12}{x + 2} (x + 2)$ $ x^2 + 2 = -9x - 12$ Subtract $-9x - 12$ from both sides: $ x^2 + 2 - (-9x - 12) = -9x - 12 - (-9x - 12)$ $ x^2 + 2 + 9x + 12 = 0$ $ x^2 + 14 + 9x = 0$ Factor the expression: $ (x + 7)(x + 2) = 0$ Therefore $x = -7$ or $x = -2$ At $x = -2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -2$, it is an extraneous solution.